xiaoxiao2021-02-28  3

/*Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.   Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).   Output For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.   Sample Input 2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5   Sample Output Case 1: 14 1 4 Case 2: 7 1 6*/ //一开始有些看不懂题意,看起来并不是求最大子序列呀 //搜了题解之后懂了,原来每一行前面要输入数组大小 //动态规划 #include<iostream> using namespace std; int main(){     int T,N,sum,start,end,max; int temp=1;     int dp;     cin>>T;     for(int k=0;k<T;k++){         cin>>N; temp=1;     /*    dp= new int[N+1];         for(int i=0;i<N;i++) cin>>dp[i];         max=dp[0];         for(start=0;start<N;start++){             sum=dp[start];             for(end=start+1;end<N;end++){                 sum+=dp[end];                 if(max<sum){//一开始我把If放在外面我发现自己每次都是重新加的,其实应该加一下判断一下                     max=sum; f=start; l=end;                 }             }         }                  delete []dp;*/         //开始想的是把start从0开始往end加,有Max就记下下标,然后输出最大的下标,时间复杂度为n^3 /结果发现超时了,毕竟是n3啊         max=-1001; sum=0; // start=1;         for(int i=0;i<N;i++){             cin>>dp;             sum+=dp;             if(sum>max){                 max=sum; start=temp; end=i+1;             }             if(sum<0){                 sum=0;如果总数比0小,说明刚刚加的数字应该舍去,所以更新temp,temp更新start                 temp=i+2;             }         }         cout<<"Case "<<k+1<<":"<<endl;         cout<<max<<" "<<start<<" "<<end<<endl;             //参考了别人的算法         if(k!=T-1) cout<<endl;     }      //如果全是负数,最大值为0,因为有一段子序列为空集     return 0; }
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